AI 中文总结
本文通过引入基于整数分拆极值问题的递归比较方法，证明了Brown和Mol关于树的子树根的猜想，刻画了等号情况并推导了非零子树根的下界，相关界渐近尖锐。
AI 中文摘要
树的子树多项式是按阶数枚举其子树的生成函数。Brown和Mol猜想，每个阶数n≥2的树的子树根都位于圆盘{z∈ℂ: |z|≤1+√[n-1]{n-1}}内。我们通过引入基于整数分拆极值问题的递归比较方法证明了该猜想，进一步刻画了等号情况：当且仅当n为偶数且该树为星型树时，上界可达，此时唯一的边界根为-1-√[n-1]{n-1}。我们还证明每个非零子树根z满足|z|>√[n-1]{n-1}-1，当n→∞时该下界渐近尖锐；对于奇数n，虽未达到上界，但当n→∞时上界渐近尖锐。
英文摘要
The subtree polynomial of a tree is the generating function that enumerates its subtrees according to their orders. Brown and Mol conjectured that every subtree root of a tree of order $n\ge 2$ lies in the disk \[ \left\{z\in\mathbb C: |z|\le 1+\sqrt[n-1]{n-1} \right\}. \] We prove this conjecture by introducing a recursive comparison method based on an extremal problem over integer compositions. We further characterize the equality case: the upper bound is attained if and only if $n$ is even and the tree is the star; in this case the unique boundary root is $-1-\sqrt[n-1]{n-1}$. We also show that every nonzero subtree root $z$ satisfies \[ |z|>\sqrt[n-1]{n-1}-1. \] The lower bound is asymptotically sharp as $n\to\infty$. For odd $n$, although the upper bound is not attained, it is asymptotically sharp as $n\to\infty$.
